math185f09-hw5sol

# math185f09-hw5sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 5 SOLUTIONS 1. Let a ∈ R and z ∈ C . (a) Evaluate the following integrals Z 1 e it cos atdt and Z 1- 1 dt t 2 + i . Solution. For the first integral, we have Z 1 e it cos atdt = Z 1 cos t cos atdt + i Z 1 sin t cos atdt. Since cos t cos at = 1 2 [cos( t + at ) + cos( t- at )] , sin t sin at = 1 2 [sin( t + at ) + sin( t- at )] , we obtain Z 1 e it cos atdt = 1 2 sin(1 + a ) t 1 + a + sin(1- a ) t 1- a 1- i 2 cos(1 + a ) t 1 + a + cos(1- a ) t 1- a 1 = 1 2 sin(1 + a ) 1 + a + sin(1- a ) 1- a- i 2 cos(1 + a ) 1 + a + cos(1- a ) 1- a- 2 1- a 2 . For the second integral, we first apply the formula for complex square roots in Chapter 1 and then find a partial fraction decomposition. 1 t 2 + i = 1 t- 1 √ 2 + i √ 2 t + 1 √ 2- i √ 2 = α t- 1 √ 2 + i √ 2 + β t + 1 √ 2- i √ 2 and so α = 1 2 √ 2 (1 + i ) , β =- 1 2 √ 2 (1 + i ) . Therefore Z 1- 1 dt t 2 + i = 1 2 √ 2 (1 + i )log t- 1 √ 2 + i √ 2 1- 1- 1 2 √ 2 (1 + i )log t + 1 √ 2- i √ 2 1- 1 . (b) Show that if Re z >- 1, then the integral R 1 t z dt exists and Z 1 t z dt ≤ 1 1 + Re z . Solution. Note that for 0 ≤ t ≤ 1, | t z | = t Re z and so Z 1 t z dt ≤ Z 1 | t z | dt = Z 1 t Re z dt. Date : October 23, 2009 (Version 1.0). 1 Since for Re z >- 1 the last integral converges, R 1 t z dt exists and the required inequality follows from Z 1 t Re z dt = t Re z +1 1 + Re z 1 = 1 1 + Re z . (c) Show that if a < 1, then Z 1- 1 cos it t a dt ≤ 2 Z 1- 1 dt t a and thus the (improper) integral R 1- 1 t- a cos itdt converges absolutely. Solution. Note that cos it = cosh t and for- 1 ≤ t ≤ 1, we have 0 ≤ cosh t < 2. Hence Z 1- 1 cos it t a dt ≤ Z 1- 1 cos it t a dt = Z 1- 1 cosh t | t a | dt ≤ 2 Z 1- 1 dt t a . Since the last integral converges for a < 1, we see that R 1- 1 t- a cos itdt converges absolutely. 2. (a) For k = 1 , 2 , 3, evaluate the following integrals Z Γ k Re( z ) dz, Z Γ k z 2 dz, Z Γ k dz z along the curves from the point z = 1 to z 1 = i in the counter clockwise direction as described in Figure 1. Figure 1. Left : Γ 1 is along the boundary of the square: { x + iy | ≤ x ≤ 1 , ≤ y ≤ 1 } . Center : Γ 2 is along the boundary of the circle: { e it | ≤ t ≤ π/ 2 } . Right : Γ 3 is along the line segment: { (1- t ) + it | ≤ t ≤ 1 } ....
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## This note was uploaded on 02/17/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at Berkeley.

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math185f09-hw5sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

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